Question

Question 1 Find the final value of x(t) with laplace transform: 53 +25 +as+b c54 +55 +25 +65 X(5)=- where a-23, b=37, and c=46.

Answer 1

#1 $+25 +as+b C5*+55°+25+65 Final value of a(t) i.e., x (0) = ? sob According to "final value theorem” in Laplace transfoom Lt act) exists OCC00) = Lt sx(s) it sto SO Com (0) = oto+b = b 0+0+8 b= 37 → 2C)= 37 = 4.625 Given 2000) = 4.625 12C) = 4625

#2. Y(5) - od sol: To solve this problem, let us go by 1 given options. @) y(t) = 4t ; t>0. Sol Y65) = 1°ych) est dt {definition of Laplace transform? As tzo for y(t) , integration lower limit starts from o to o Y(s) = soft est at = 45 t est at

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Y(s) = 4 pr est at But Y(s) should be a so option (a) is wrong. Ilu b) y(t) = 4_u(t) sol By defition of Laplace transform 44 of -sh Y(3) - S4 et dt fi step function som L wth starts from =45&st at too} +43*** Option (b) is correct one:

C) y(t) = 247 ;t20 sel By definition of Laplace transform Y(6) - Jett est at gylt) - jestest at at Here also ylt) +4) titooo starts ſ from t=0 = e(+4) tyto лие, | Y) = 1 so option (C) is not true Cel) y(t) = 48(t) = 5 4; t=0, ? We know that 20; else where 14 O 18 (1) dx=1. lie, Area under impulse function is 1 Y(S) = 4584e* dt + >[YC) = 4) :option (d) is also woong. woong

o-joo Conclusion #1. Option (b) is correct #2. Directly applying Inverse from Laplace transform also gets the same answer. '{FC+} = Pcs) est ds. Nole points ; #3. Final value theorem is of Laplace transform is valid if all poles have negative real parts except a simple pole at origin #4. Poles in jw-axis con right half of s-plane ; final value theorem is not applicable